SOLUTION: 16x^2-24x-13=0 solve by the method of completing the squares

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Question 696758: 16x^2-24x-13=0 solve by the method of completing the squares
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+16x%5E2+-+24x-13=0+
+16x%5E2+-+24x+=+13+
Divide both sides by +16+
+x%5E2+-+%283%2F2%29%2Ax+=+13%2F16+
Take 1/2 of the co-efficient of the +x+ term,
square it, then add it to both sides
+x%5E2+-+%283%2F2%29%2Ax+%2B+%28+-3%2F4+%29%5E2+=+13%2F16+%2B+%28+-3%2F4+%29%5E2+
+x%5E2+-+%283%2F2%29%2Ax+%2B+%28+-3%2F4+%29%5E2+=+13%2F16+%2B+%28+-3%2F4+%29%5E2+
+x%5E2+-+%283%2F2%29%2Ax+%2B+9%2F16+=+13%2F16+%2B+9%2F16+
+x%5E2+-+%283%2F2%29%2Ax+%2B+9%2F16+=+22%2F16+
+%28+x+-+3%2F4+%29%5E2+=+22%2F%284%5E2%29+
Take the square root of both sides. The left side is a perfect square
+x+-+3%2F4+=+sqrt%2822%29+%2F+4+
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The solutions are:
+x+=+%28+3+%2B+sqrt%2822%29+%29+%2F+4+
+x+=+%28+3+-+sqrt%2822%29+%29+%2F+4+
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check the answers:
+x+=+%28+3+%2B+sqrt%2822%29+%29+%2F+4+
+x%5E2+=+%28+9+%2B+6%2Asqrt%2822%29+%2B+22+%29+%2F+16+
-----------------------
+16x%5E2+-+24x+-13+=+0+

+9+%2B+6%2Asqrt%2822%29+%2B+22++-+6%2A%28+3+%2B+sqrt%2822%29+%29+-+13+=+0+
+9+%2B+22+-+18+-+13+%2B+6%2Asqrt%2822%29+-+6%2Asqrt%2822%29++=+0+
+31+-+31+%2B+6%2Asqrt%2822%29+-+6%2Asqrt%2822%29++=+0+
+0+=+0+
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You can check the other solution